f(x)=xsinx-3/2
f'(x)=sinx+xcosx
令f'(x)=0得tanx=-x,解为x0,x0∈(π/2,π)
∴(0,x0),f(x)递增,（x0,π),f(x)递减
f(x)max=f(x0)>f(π/2)>0
f(0)=-3/2